【Codeforces】-632B-Alice, Bob, Two Teams（模拟，思维）

最新推荐文章于 2020-02-03 15:49:07 发布

Bear1997

最新推荐文章于 2020-02-03 15:49:07 发布

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分类专栏：模拟思维

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在一场游戏中，Alice和Bob需要将游戏物品分成两个团队。Bob需要根据Alice初始分配的团队来确定最佳策略以最大化他的团队实力。给定物品的强度和分配，Bob可以选择前缀或后缀翻转团队标识。目标是找到能最大化Bob团队实力的策略。题解涉及扫描线算法，计算不变和变化时B团队的实力。

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Alice and Bob are playing a game. The game involves splitting up game pieces into two teams. There are n pieces, and the i-th piece has a strength pi.

The way to split up game pieces is split into several steps:

First, Alice will split the pieces into two different groups A and B. This can be seen as writing the assignment of teams of a piece in an n character string, where each character is A or B.
Bob will then choose an arbitrary prefix or suffix of the string, and flip each character in that suffix (i.e. change A to B and B to A). He can do this step at most once.
Alice will get all the pieces marked A and Bob will get all the pieces marked B.

The strength of a player is then the sum of strengths of the pieces in the group.

Given Alice's initial split into two teams, help Bob determine an optimal strategy. Return the maximum strength he can achieve.

Input

The first line contains integer n (1 ≤ n ≤ 5·105) — the number of game pieces.

The second line contains n integers pi (1 ≤ pi ≤ 109) — the strength of the i-th piece.

The third line contains n characters A or B — the assignment of teams after the first step (after Alice's step).

Output

Print the only integer a — the maximum strength Bob can achieve.

Examples

input
5
1 2 3 4 5
ABABA

output
11

input
5
1 2 3 4 5
AAAAA

output
15

input
1
1
B

output
1

Note

In the first sample Bob should flip the suffix of length one.

In the second sample Bob should flip the prefix or the suffix (here it is the same) of length 5.

In the third sample Bob should do nothing.

题意：：题目是说从前或从后选一个子串（必须包括首或者尾，长度任意），把A变B，B变A，然后计算B的总价值，使之最大。

题解：先统计不变时B的价值，然后从前从后各扫一遍就行了

sum1记录从前到后变化A B，值的变化。sum2记录从后到前变化......

如：从前到后扫一遍，i=3,那么sum1就继续改变(i=1,i=2 .A变成B，B变成A已经记录)，再更新ans

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
int a[500010];
char s[500010];
int main()
{
	int n;	
	while(~scanf("%d",&n)) 
	{
		__int64 ans=0,sum1,sum2;
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);	
		scanf("%s",s+1);
		for(int i=1;i<=n;i++)
		{
			if(s[i]=='B')
				ans+=a[i];
		}
		sum1=ans;
		sum2=ans;
		for(int i=1;i<=n;i++)	//从前到后一个for循环,两个for会超时.要包括首以及i之前的字母那么sum1就一直更新就好 
		{
			if(s[i]=='A')		
				sum1+=a[i];
			else
				sum1-=a[i];
			ans=max(ans,sum1);
		}
		for(int i=n;i>=1;i--)
		{
			if(s[i]=='A')
				sum2+=a[i];
			else
				sum2-=a[i];
			ans=max(ans,sum2);
		}
		printf("%I64d\n",ans);
	}
	return 0;
}